RPGWW

You all know about factoring, right? Y'know, the way you turn x^2 -7x+12 into (x-3)(x-4)?

Well, got one here for you to bust your brains over. Be warned, it involves that kooky little imaginary number, i (the square root of -1).

FACTOR OUT:
13i + 18


Good luck, simpletons. <p>--
<div style="text-align:center">This has been your daily interrupted moment of Zen.

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I really don't see why math belongs in the spam forum could you please maybe mosey on elsewhere with your sorcery?

.... but it's not a variable!

It's like factoring 5 + 2^(1/2)!

Lys: It functions as a variable because it's an imaginary number. But a little different. It can still work for factoring purposes.

Also: I will attempt this problem if Xaq will attempt the most elementary and simple task of using the concept of "i" to factor the following:

-5

... This isn't a joke either. It can be done, but I'm tired and have spent far too long looking at this problem.

Edit: THE FIVE WAS ALWAYS NEGATIVE. ... REALLY. <p>

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<div style="text-align:center"></div></p>Edited by: Besyanteo&nbsp; at: 11/18/05 14:07

Lys: (ai+b)(ci+d) = aci^2 + (bc+ad)i + bd = ac(-1) + (bc+ad)i + bd = (bc+ad)i + (bd-ac)

That help to see what Will means by factoring it?

Edit: Doing a quick bit of linear algebra, I found a general form for finding such a value for any complex number, but I don't want to ruin it for everyone else. Could you IM me so I can confirm my answer? :D

Edit2: I couldn't wait. :D

Don't click here if you don't want to be spoiled. www.mysidia.org/nanimono/solution.txt

Edit3: Thinking about it, this isn't just how to factor a complex number into two other complex numbers, it's also instructions on how you would divide one complex number ei+f by another complex number ai+b. Which makes sense, in a way, given the sense of factoring meaning purely dividing something up into its factors. Like the factors of 6 are 2 and 3, or restated, 6 divided by 2 is 3. Math is awesome! <p>

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"Never let your morals get in the way of doing what is right" - Salvor Hardin
</p>Edited by: Idran1701 at: 11/18/05 10:48

Awesomeness! I learn new things everyday!

I haven't seen i since high school, and I am glad of this.

I much prefer statistics, sad as that sounds, to stuff like i. <p><div style="text-align:center"></div></p>

i? Quite frankly, that sounds made up. <p>

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18+13i = (4+i)(5+2i) and both factors are Gaussian primes. :P <p>-----
Do not taunt Happy Fun Zemyla.

<span style="font-size:xx-small;">I think boobs are the lesser of two evils. - Inverse (Pervy)
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Ultimately, wizards and clerics don't say, "Gee, I want to become a lich because weapons hurt less and I don't have to worry about being backstabbed; that whole 'eternal life' thing is just a fringe benefit."-Darklion
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</span>

Adventurers! | RPG World World | The Phantom Lord's OT Board mkII | Indie Madnesse | Brotherhood of Elitist Bastards</p>

A complex number ei + f can be divided into two factors: ai + b, where a and b are any two real numbers where a and b are not both
equal to zero, and ci + d, where c = (eb - fa)/(a^2 + b^2) and d = (ea + fb)/(a^2 + b^2)

13i + 18

e=13
f=18

c = (-18a + 13b)/(a^2 + b^2)
d = (13a + 18b)/(a^2 + b^2)

(ai+b)(ci-d)

a=2
b=0

c = (-18(2) + 13(0))/((2)^2 + (0)^2)
c = (-36)/(4)
c = -9
d = (13(2) + 18(0))/((2)^2 + (0)^2)
d = (26)/4
d = 6.5

(2i+0)(-9i-6.5)
-(18i^2)+13i+0+0
-(18*-1)+13i
-(-18)+13i
18+13i
13i+18

... Well hot damn. They said any two real numbers. I didn't think they meant you could just arbitrarily pick them.

Also, becasue this all became much easier:

-5
(2i-1)(2i+1)
4i^2-2i+2i-1
(4*-1)-1
-4-1
-5

How did the previous question become positive?! ... It was almost 6 in the morning. Typos do occur. Bite me.

Edit: Zem's answer is superior, because it has no decimals and is therefore, you know, legal. <p>

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<div style="text-align:center"></div></p>Edited by: Besyanteo&nbsp; at: 11/18/05 14:06

It's factorable though I have to wonder what the point to factoring that is. With variables, the act of factoring is part of a process of discovering values for the variables or any other number of wonderful things involved in the equation/statement. With i, what's the point? It could just be that my math is deficient in this sense, but it seems like an exercise in happy fun pointlessness. <p>[---------------------------]
"There is great disorder under Heaven, and the situation is excellent."
-Mao Tse-Tung

"I eat the talking bees because I am George Washington Christ"
-From "Bob the Ball"</p>

Factoring in the gaussian numbers isn't all that hard. The first thing you need to do, though, is factor a*a+b*b first, due to the properties of complex multiplication. That will give you the magnitudes of the necessary complex numbers. After that (since there are at least 8 gaussian integers with the same amplitude, but most only differ by a factor of ±1 or ±i) just combine the various numbers with the same amplitude until you get the result. <p>-----
Do not taunt Happy Fun Zemyla.

<span style="font-size:xx-small;">I think boobs are the lesser of two evils. - Inverse (Pervy)
Dammit, Dan, I'm not dating a damn NPC! - OOC Will (Will Baseton)
Of course! Anything worth doing is worth doing completely wrong! - Travis English
Ultimately, wizards and clerics don't say, "Gee, I want to become a lich because weapons hurt less and I don't have to worry about being backstabbed; that whole 'eternal life' thing is just a fringe benefit."-Darklion
But this one time I killed a walrus with my bare hands, and I suddenly understood spherical coordinates. - KnightsofSquare
Also, when you've worked a 36-hour shift as an intern you too just might pour yourself a catful of coffee and sit down to cuddle with your travel mug. -eirehound
</span>

Adventurers! | RPG World World | The Phantom Lord's OT Board mkII | Indie Madnesse | Brotherhood of Elitist Bastards</p>

Belatedly: Figured out how I got 5 again, since getting on the proper computer. I mixed up my dern coefficients. :{

(2+i)(2-i)
4-2i+21-(-1)
5

Also: I thick we solved Xaq's question and stuff. I'm totally going to stop posting now. And maybe take a nap. <p>

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I have no idea what a Gaussian prime is. I just found my method by taking the equation (bc+ad)i + (bd-ac) = ei + f, plugging it in as a system of two equations bc+ad = e and bd-ac = f, and solving it as a 2x3 augmented matrix

{ b a e}
{-a b f}


I thought I found something nifty. :( <p>

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"Never let your morals get in the way of doing what is right" - Salvor Hardin
</p>Edited by: [url=http://p068.ezboard.com/brpgww60462.showUserPublicProfile?gid=idran1701>Idran1701</A] at: 11/18/05 19:38

Idran: You did. Honestly. This is going to help me lots in Algebra 2, with exam time coming up. <p>

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Oh. Awesome! Glad I could be unintentionally helpful. <p>

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"Never let your morals get in the way of doing what is right" - Salvor Hardin
</p>